Download Binomial Probability Distribution

Binomial Probability
Distribution
Animal Health Risk Analysis
Canadian Food Inspection Agency
3851 Fallowfield Road
Ottawa, Ontario, Canada K2H 8P9
Canadian Food
Inspection Agency
(CFIA)
Analyse des risques de la santé des animaux
Agence canadienne d’inspection des aliments
3851, chemin Fallowfield
Ottawa (Ontario) Canada K2H 8P9
Agence canadienne
d’inspection des aliments
(ACIA)
Properties of the Binomial Distribution
One of the most elementary discrete random variables is the coin-tossing
experiment. Each toss is called a trial. In any single trial their will be a probality
associated with a particular event such as a head on the coin. The probability will
not change from one trial to the next. Such trials are said to be independent and
are often called Bernoulli trials.
Bernoulli trials
A Bernoulli trial has only two outcomes: success or failure, denoted as 1 or 0. If
the probability of success is p the Bernoulli distribution is f(x) = px q1-x.
Example: Coin toss with a balanced coin, where 0 denotes tail and 1 denotes head.
When x = 0, f(x) = 0.50 0.51-0 = 1 × ½ =½ and when x = 1, f(x) = 0.51 0.51-1 = ½ × 1 = ½.
Binomial Distribution
A binomial probability distribution determines the probabilities for the number of
wins (successes) in n identical trials where these conditions are met:
n identical trials
each trial consists of two outcomes, either a success, S, or a failure, F
the probability of success on a single trial is equal to p and remains the same
from trial
to trial, the probability of failure is equal to (1-p) = q
Binomial Probability Distribution
CFIA/ACIA
Properties of the Binomial Distribution
Numerous coin tossing experiments are conducted almost daily in regulatory and
academic veterinary medicine. A survey of swine producers to determine if they
feed household food waste to their pigs is one such experiment. Each swine
producer selected is analagous to the toss of an unbalanced coin, since the
probability of a ‘yes’ is not one-half but very likely a small probability. Other
experiments include: studies to determine the effectiveness of a La Sota Newcastle
disease virus vaccine in chickens to protect against experimental challenge with
velogenic virus; a survey of cattle ranchers to estimate the compliance with FMD
vaccination of all cattle over 4 months of age; and a study to estimate the
proportion of longissimus dorsi muscles which have a pH value of less than 6.2 at
24 hours of maturation. These experiments approximate for all practical purposes
binomial experiments. Each trial will result in one of two outcomes (protection or
not; compliance or not; and pH above or below 6.2). The probability for success
will remain the same for trial to trial and will be unaffected by the outcome on any
of the other trials. In the survey of the cattle ranchers, the probability for success
will remain approximately constant as long as the population of cattle ranchers is
large in comparison with the survey sample n.
n = a fixed number of identical Bernoulli trials
p = the probability of success in each trial
X = the number of successes in n trials
Binomial Probability Distribution
CFIA/ACIA
Properties of the Binomial Distribution
The random variable X is called a binomial random variable. Its distribution is called
a binomial distribution. The probability of x successes in n trials is:
P( X = x ) = C xn p x q n − x =
n!
p x q n− x
x !(n − x )!
Where
C xn =
n!
x !( n − x )!
For values of x = 1, 2, ..., n and where n! =n(n-1)(n-2) ... (2)(1) and 0! = 1.
Rule of Thumb
When the sample (the n identical trials) comes from a large population, the
probability of success p is the same from trial to trial. When the population size N is
small, the probability of success p can change dramatically from trial to trial and the
experiment is not binomial.
If the sample size is large relative to the population size, that is, n/N$0.05, then the
resulting experiment is not binomial.
Binomial Probability Distribution
CFIA/ACIA
Properties of the Binomial Distribution
Probability density function:
 n
f ( x ) =   p x (1 − p) n − x
 x
Cumulative distribution function:
 n i
F ( x) = ∑   p (1 − p) n− i
i=0  i 
x
Parameter: n>0, n is an integer, 0#p#1
Domain: {0, 1, 2, 3, ..., n}
Mean (:): np
Variance (F2): npq
Mode: p (n+1)-1 and p(n+1) if p(n+1) is an integer, lp(n+1)m otherwise
Coefficient of skewness ("3): (q-p)/%&npq
l m = Greatest integer function
Coefficient of kurtosis ("4): 3 - 6/n + 1/npq
Binomial Probability Distribution
CFIA/ACIA
Properties of the Binomial Distribution
0.388
Probability mass
function for BIN(n = 10,
p = 0.1)
0.310
0.233
0.155
0.078
0.000
0
1
2
3
4
0
1
2
3
4
1.000
Cumulative
distribution function
for BIN(n = 10, p = 0.1)
0.800
0.600
0.400
0.200
0.000
Binomial Probability Distribution
CFIA/ACIA
Binomial Cumulative Probabilities Table Examples
What is the probability that X#2 for a binomial random variable with n = 5 and p =
0.2?
This can be written P(BIN(n =5, p = 0.2) # 2) = F(2)
2
=
 5
∑  x 0.2
x
× 0.85− x = 0.9421
x=0
For P(2#BIN(n =5, p = 0.2) # 4) = F(4) - F(2) =P(X#4) - P(X#2) = 0.9997- 0.9421 = 0.0576
from the binomial cumulative probabilities tables on the next slide. The cumulative
probabilities are obtained from the following expression:
 n
P( X ≤ x) = ∑   p i (1 − p) n−i
i=0  i 
x
Binomial Probability Distribution
CFIA/ACIA
Binomial Cumulative Probabilities Table Examples
Table of Binomial Distribution Cumulative Probabilities
p
n
x
2
3
4
5
0
1
2
0
1
2
3
0
1
2
3
4
0
1
2
3
4
5
0.05
0.9025
0.9975
1.0000
0.8574
0.9928
0.9999
1.0000
0.8145
0.9860
0.9995
1.0000
1.0000
0.7738
0.9774
0.9988
1.0000
1.0000
1.0000
0.1
0.8100
0.9900
1.0000
0.7290
0.9720
0.9990
1.0000
0.6561
0.9477
0.9963
0.9999
1.0000
0.5905
0.9185
0.9914
0.9995
1.0000
1.0000
Binomial Probability Distribution
0.15
0.7225
0.9775
1.0000
0.6141
0.9393
0.9966
1.0000
0.5220
0.8905
0.9880
0.9995
1.0000
0.4437
0.8352
0.9734
0.9978
0.9999
1.0000
0.2
0.6400
0.9600
1.0000
0.5120
0.8960
0.9920
1.0000
0.4096
0.8192
0.9728
0.9984
1.0000
0.3277
0.7373
0.9421
0.9933
0.9997
1.0000
0.25
0.5625
0.9375
1.0000
0.4219
0.8438
0.9844
1.0000
0.3164
0.7383
0.9492
0.9961
1.0000
0.2373
0.6328
0.8965
0.9844
0.9990
1.0000
0.3
0.4900
0.9100
1.0000
0.3430
0.7840
0.9730
1.0000
0.2401
0.6517
0.9163
0.9919
1.0000
0.1681
0.5282
0.8369
0.9692
0.9976
1.0000
0.35
0.4225
0.8775
1.0000
0.2746
0.7183
0.9571
1.0000
0.1785
0.5630
0.8735
0.9850
1.0000
0.1160
0.4284
0.7648
0.9460
0.9947
1.0000
0.4
0.3600
0.8400
1.0000
0.2160
0.6480
0.9360
1.0000
0.1296
0.4752
0.8208
0.9744
1.0000
0.0778
0.3370
0.6826
0.9130
0.9898
1.0000
0.45
0.3025
0.7975
1.0000
0.1664
0.5748
0.9089
1.0000
0.0915
0.3910
0.7585
0.9590
1.0000
0.0503
0.2562
0.5931
0.8688
0.9815
1.0000
0.5
0.2500
0.7500
1.0000
0.1250
0.5000
0.8750
1.0000
0.0625
0.3125
0.6875
0.9375
1.0000
0.0313
0.1875
0.5000
0.8125
0.9688
1.0000
CFIA/ACIA
Limousine Survival Example
Ten years ago 5 purebred Limousine heifer calves were imported for 5 different owners
from a country subsequently reporting cases of bovine spongiform encephalopathy. An
effort to find these imported cattle was about to be initiated after first answering the
question of the probability that all 5 would be alive and the probability that at least one
would be alive. Based on national statistics of survival of purebred beef cattle at 10
years of age, the probability of survival to 10 years of age was 0.25.
These questions involve the Binomial distribution because 1) the experiment consists of
5 identical trials, 2) each trial results in one of two outcomes, 3) the probability of
success on single trial is equal to p = 0.25, and 4) the trials are independent. The
probability that all 5 cattle would be alive represents the case in which 5 successes out
of 5 trials is entertained.
P( X = x ) =
5!
n!
0.255 × 0.750 = 0.001
p x q n− x =
5!(0)!
x !(n − x )!
The probability that at least one cow would be living is estimated as follows:
P( X ≥ 1) = P( X = 1) + P( X = 2) + P( X = 3) + P( X = 4) + P( X = 5)
P( X ≥ 1) = 1 − P( X < 1) = 1 − P( X = 0) = 1 −
5!
× 0.250 × 0.755 = 1 − 0.24 = 0.76
0!(5 − 0)!
Binomial Probability Distribution
CFIA/ACIA
Limousine Survival Example
Another approach is to generate probability density function and cumulative distribution
function tables for the binomial distribution where n = 5 and p = 0.25. Such tables can
be created in Microsoft Excel using the following BINOMDIST function for this example,
employing a 0 for the last parameter to represent the probability density function and 1
for the cumulative distribution function.
Probability density function
giving the individual
binomial probabilities for
BIN(5,0.25).
Cumulative distribution
function giving the
cumulative binomial
probabilities for
BIN(5,0.25).
x
0
1
2
3
4
5
P(X=x)
0.237305
0.395508
0.263672
0.087891
0.014648
0.000977
0
1
2
3
4
5
P(X<=x)
0.237305
0.632813
0.896484
0.984375
0.999023
1
x
=BINOMDIST(0,5,0.25,0)
=BINOMDIST(0,5,0.25,1)
Both questions P(X=5) and P(X$1) = 1 - P(X=0) can be answered from the above tables.
Binomial Probability Distribution
CFIA/ACIA
Relationship Between Binomial and Normal Probability
Distributions
Probabilities associated with binomial experiments are readily obtainable for the
formula BIN(n,p) of the binomial distribution or from binomial tables when n is
small. The normal distribution is often a good approximation to a discrete
distribution when the latter takes on a symmetric bell shape. The binomial
distribution is nicely approximated by the normal in practical problems when one
works with the cumulative distribution function.
If X is a binomial random variable as the number of successes in n independent
trials with p as the probability of success and 1-p as the probability of failure on
any single trial and with mean (:) =np and variance F2 = npq, then the limiting form
of the distribution of:
Z=
X − np
npq
as n 64, is the standard normal distribution N(0, 1) or NORMALZ(0,1).
It turns out that the normal distribution with : = np and F2 = np(1-p) not only
provides a very accurate approximation to the binomial distribution when n is large
and p is not extremely close to 0 or 1, but also provides a fairly good approximation
when n is small and p is reasonably close to ½.
Binomial Probability Distribution
CFIA/ACIA
Relationship Between Binomial and Normal Probability
Distributions
Note:
Since we are approximating a discrete random variable with a continuous
one, we need to apply a continuity correction. That is, if a binomial random variable
X takes on the value ‘a’ exactly, then the corresponding normal X must take on the
interval of values (a - 0.5, a + 0.5) since the probability of an exact value for a
continuous randomvariable is always 0.
Example - The rate of operative complications in a certain complex vascular
reconstructive procedure is 30%. This includes all complications, ranging in
severity from wound separation or infection to death, and is the proportion of
patients showing some complication. Assuming independence of the occurrence
of complication in different patients, in a series of 100 such procedures, what is the
probability that there will be:
(a) at most 18 patients with operative complication?
(b) exactly 16 patients with operative complication?
Let X = # of complications occurring.
(a)
 100
 (.3) x (.7) 100− x
x 
x= 0
18
P( X ≤ 18) =
∑ 
Binomial Probability Distribution
CFIA/ACIA
Relationship Between Binomial and Normal Probability
Distributions
We need to use the normal approximation since the probability cannot be found in
the binomial tables. If X' is now the corresponding r.v. we want P(XN# 18.5).
Q P( X ' ≤ 18.5) = P( Z ≤
= P( Z ≤
(b)
18.5 − nπ
nπ (1 − π
18.5 − 30
) ≅ P( Z ≤ − 2.51) = 0.006
21
P( X = 16) = P(155
. ≤ X ' ≤ 16.5) = P(
155
. − 30
16.5 − 30
≤ Z≤
)
21
21
≅ P( − 316
. ≤ Z ≤ − 2.95) = P( Z ≤ − 2.95) − P( Z ≤ − 316
. )
= 0.0016 − 0.0008 = 0.0008
Binomial Probability Distribution
CFIA/ACIA
Relationship Between Binomial and Normal Probability
Distributions
Example - Find the probability that a student can guess from 110 to 120 correct
answers out of 200 questions on a true-false examination. (No subtraction of wrong
from right is done.)
Let X = no. of correct answers out of 200 questions on an exam.
ˆ X is binomial with n = 200 and B = ½.
ˆ P(110#X#120) =P(109.5#XN#120.5) if X is N(nB, nB (1-B))
= P(
109.5 − 100
120.5 − 100
≤ Z≤
) ≅ P(134
. ≤ Z ≤ 2.90)
50
50
= P( Z ≤ 2.90) − P( Z ≤ 134
. ) = .9981− .9099 = .0882
ˆ a student will guess 110 to 120 correct answers out of the 200 T-F exam questions
with an 8.8% chance.
Note:
If in a binomial distribution n is very large while B is very small, let m =
nB and use the Poisson probability
n
c x π x (1 − π ) n − x
for any x value.
Binomial Probability Distribution
e −mm x
to approximate the binomial probability
x!
CFIA/ACIA